 So a binomial is the sum, or difference, of two terms. x plus 3, 6 minus z, a plus b times t plus d, or 7 plus 3x plus 5y. And importantly, sometimes what we call a binomial may have other things involved in it, but remember the last operation performed defines a type of expression. So here the last thing we do is we add, and so these are the sum of two terms. And many problems require finding one or more terms of the power of a binomial. So let's suppose we wanted to find the x cubed term in the expansion of 2x plus 7 raised to power 5. Well, we could just multiply it out. But that's a lot of work, and so we'd rather not do that. What about our menu choices? Well, since there are five factors, we choose a term from each factor and multiply them together. So if I want an x cubed term, well, let's see. All of our factors look alike, so maybe I'll choose the 2x term for the first factor, the 2x term for the second factor, the 2x term for the third factor. That gets me an x cubed, so I need to choose the seven terms from the fourth and fifth factors. But we have to choose every possible way. So maybe we'll choose the 2x term for the first factor, the 2x term for the second factor, the seven term for the third factor, the 2x term for the fourth factor, and the seven term for the fifth factor. Or maybe I'll choose 2x, 7, 2x, 2x, and 7. Or, well, there's a lot of ways we could get an x cubed term, and so we'd rather not do it this way because it's very tedious, and unless we're very careful, we'll miss terms. So let's shift a few points a little bit. Let's consider a different approach. To get the x cubed term, what we need to do is to choose the 2x from three of the factors and then choose the seven from the remaining factors. Equivalently, we might just say choose three of the factors to provide a 2x, and the remaining factors will provide a 7. And so the question we want to ask is, how many ways can we do this? This leads us to a new branch of mathematics known as combinatorics. Combinatorics is the branch of mathematics that asks the question, how many? And a key result in combinatorics is what's known as the binomial coefficient. Suppose you want to pick k objects from a set of n objects. The number of ways you can do this is given by a formula. Over here on the left-hand side, we typically read this as n choose k, and we compute it by the formula on the right-hand side, where the exclamation point here is read factorial. So this is n factorial, where m factorial is the product of the whole numbers from 1 to m, and every now and then we'll have to deal with 0 factorial, and that's defined to be 1. So for example suppose you have 12 donuts. How many different sets of 5 can you choose? So here we're picking 5 objects from a set of 12 objects. And this is exactly the type of thing we can compute using our binomial coefficient, so let's set that up. That's 12, choose 5, and our binomial theorem says that that's going to be 12 factorial divided by 5 factorial times 12 minus 5 factorial. And we can at least do 12 minus 5 to simplify. Now the factorials indicate the products of the whole numbers from 1 up to n. So we could calculate 12 factorial, well that's the product 1 times 2 times 3 all the way up to 12, 5 factorial, that's the product 1 times 2 all the way up to 5, and 7 factorial, that's this product. Let's take advantage of the fact that most of these factors are the same, and we're working with a fraction, so many of them will factor out. So paper is cheap, let's at least write it down. 12 factorial, 5 factorial, and 7 factorial, well our 12, choose 5 will look like this. And for reasons that will become apparent in a second, we'll remove this set of common factors from numerator and denominator, and we'll simplify, our denominator will be the products 1 through 5, and the factors that are left in our numerator will write them in descending order from 12 down to 8, and we can compute this value. Now you might wonder why we simplified the way we did. Now the reason we did this is that this allows us an easy way to remember how to calculate these binomial coefficients. First of all, notice that our denominator is the numbers from 1 through 5, and 5 is the number of things we're choosing. Next our numerator is the numbers from 12 down to 8, and 12 is the number that we're choosing from, and the important thing to notice here is that not only are these top numbers in descending order, there are exactly 5 of them. The number of factors in the numerator is exactly equal to the number of factors in the denominator, and so one way to simplify n, choose k is to have the denominator k factorial and the numerator the k descending numbers from n. Well let's see if we can calculate some terms. So let's find the m cubed f squared term in the expansion m plus f to the fifth. Now let's think about this. To get a m cubed f squared term, we have to choose 3 of the factors to be m, and 2 of the factors to be f. But since we have to choose 5 factors, we can just choose which 3 will be m, and the rest will be f. Or we could also choose which 2 will be f, and allow the rest to be m. So we can choose 3 factors of m from the 5 available factors, that's 5 choose 3, or we can choose 2 factors of f from the 5 available factors in 5 choose 2 ways. But which one is correct? Well let's compute. 5 choose 3, well that's 5 factorial divided by 3 factorial times 5 minus 3 factorial. I can find 5 minus 3, that's 2. 5 factorial is the product of the numbers from 1 through 5. 3 factorial is the product of the numbers from 1 through 3. And 2 factorial is the product of the numbers from 1 to 2. We'll remove the common factors and simplify. What about 5 choose 2? Well let's compute that. 5 factorial divided by 2 factorial times 5 minus 2 factorial. That's 5 factorial divided by 2 factorial 3 factorial. Writing these out, removing the common factors, and computing. And now we have to decide which of these two answers is the correct answer. Well actually they're the same answer, and since they're the same we can use either one. So there are 10 ways to get a term that includes 3 factors of m and 2 factors of f. And so this term will be 10 m cubed f squared.